BANACH SPACES WITH THE 2-SUMMING PROPERTY

X has the 2-summing property if the norm of every linear operator from X to a Hilbert space is equal to the 2-summing norm of the operator. Up to a point, the theory of spaces which have this property is independent of the scalar eld: the property is self-dual and any space with the property is a nite dimensional space of maximal distance to the Hilbert space of the same dimension. In the case of real scalars only the real line and real `2 have the 2-summing property. 1 In the complex case there are more examples; e.g., all subspaces of complex `3 and 1 their duals.Abstract. A Banach space

Some important classical Banach spaces; in particular, C(K) spaces, L1 spaces, the disk algebra; as well as some other spaces (such as quotients of L1 spaces by re exive subspaces K], Pi]), have the property that every (bounded, linear) operator from the space into a Hilbert space is 2-summing. (Later we review equivalent formulations of the de nition of 2-summing operator. Here we mention only that an operator T : X ! `2 is 2-summing provided that for all operators u : `2 ! X the composition Tu is a Hilbert-Schmidt operator; moreover, the 2-summing norm 2 (T) of T is the supremum of the Hilbert-Schmidt norm of Tu as u ranges over all norm one operators u : `2 ! X.) In this paper we investigate the isometric version of this property: say that a Banach space X has the 2-summing property provided that 2(T) = kT k for all operators T : X ! `2 . While the 2-summing property is a purely Banach space concept and our investigation lies purely in the realm of Banach space theory, part of the motivation for studying the 2-summing property comes from operator spaces. In Pa], Paulsen de nes for a Banach space X the parameter (X) to be the supremum of the completely bounded norm of T as T ranges over all norm one operators from X into the space B(`2 ) of all bounded linear operators on `2 and asks which spaces X have the property that (X) = 1. Paulsen's problem and study of (X) is motivated by old results of von Neumann, Sz.-Nagy, Arveson, and Parrott as well as more recent research of Misra and Sastry. The connection between Paulsen's problem and the present paper is Blecher's result B] that (X) = 1 implies that X has the 2-summing property. Another connection is through the property (P) introduced by1991 Mathematics Subject Classi cation. Primary 46B07 Secondary 47A67, 52A10, 52A15. The rst author was supported as Young Investigator, NSF DMS 89-21369; and NSF DMS 93-21369; the second and fourth authors were Workshop participants, NSF DMS 89-21369; the third author was supported by NSF DMS 90-03550 and 93-06376; the third and fourth authors were supported by the U.S.-Israel Binational Science Foundation. Typeset by AMS-TEX 1

0. Introduction:

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A. ARIAS, T. FIGIEL, W. B. JOHNSON AND G. SCHECHTMAN

Bagchi and Misra BM], which Pisier noticed is equivalent to the 2-summing property. However, since we shall not investigate here (X) or property (P) directly and do not require results from operator theory, we refer the interested reader to Pa] and BM] for de nitions, history, and references. On the other hand, since the topic we treat here is relevant for operator theorists, we repeat standard background in Banach space theory used herein for their bene t. In Pa] Paulsen asks whether (X) = 1 only if X is a one or two dimensional C(K) or L1 space; in other words, ignoring the trivial one dimensional case, whether (X) = 1 implies that X is two dimensional, and among two dimensional spaces, whether only `2 and `2 satisfy this identity. He proves that (X) = 1 implies 1 1 that dim(X) is at most 4, that (X) = (X ), and he gives another proof of Haagerup's theorem that (`2 ) = 1. Paulsen, interested in operator theory, is 1 referring to complex Banach spaces, so `2 is not the same space as `2 . 1 1 From the point of view of Banach space theory, it is natural to ask which Banach spaces have the 2-summing property both in the real and the complex cases, and here we investigate both questions. Up to a point, the theory is independent of the scalar eld: In section 2 we show that the 2-summing property is self-dual, that only spaces of su ciently small ( nite) dimension can have the property, and that a space with the property is a maximal distance space{that is, it has maximal Banach-Mazur distance to the Hilbert space of the same dimension. The main result in section 2, Proposition 2.6, gives a useful condition for checking whether a space has the 2-summing property which takes a particularly simple form when the space is 2-dimensional (Corollary 2.7.a). The analysis in section 3 yields that the situation is very simple in the case of real scalars; namely, IR and `2 are the only spaces which have the 2-summing 1 property. Two ingredients for proving this are Proposition 3.1, which says that there are many norm one operators from real `3 into `2 which have 2-summing 1 2 norm larger than one, and a geometrical argument, which together with a recent lemma of Maurey implies that a maximal distance real space of dimension at least three has a two dimensional quotient whose unit ball is a regular hexagon. The complex case is a priori more complicated, since `2 and `2 both have the 1 1 2-summing property but are not isometrically isomorphic. In fact, in section 4 we show that there are many other examples of complex spaces which have the 2-summing property; in particular, `3 and all of its subspaces. The simplest way 1 to prove that these spaces have the 2-summing property is to apply Proposition 2.6, but we also give direct proofs for `3 in section 4 and for its two dimensional 1 subspaces in the appendix. The case of `3 itself reduces via a simple but slightly 1 strange \abstract nonsense" argument to a calculus lemma, which, while easy, does not look familiar. (In BM] the authors give an argument that `3 satis es their 1 property (P) which uses a variation of the calculus lemma but replaces the \abstract nonsense" with a reduction to self-adjoint matrices.) We also give in Proposition 4.5 an inequality which is equivalent to the assertion that all two dimensional subspaces of complex `3 have the 2-summing property. While we do not see a simple direct 1 proof of this inequality, we give a very simple proof of a weaker inequality which is equivalent to the assertion that every two dimensional subspace of complex `3 is 1 of maximal distance. In section 5 we make some additional observations.

BANACH SPACES WITH THE 2-SUMMING PROPERTY

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1. Preliminaries.

Standard Banach space theory language and results can be found in LT1], LT2], while basic results in the local theory of Banach spaces are contained in T-J2]. However, we recall here that part of the theory and language which we think may not be well known to specialists in operator theory. Spaces are always Banach spaces, and subspaces are assumed to be closed. Operators are always bounded and linear. The Banach-Mazur] distance between spaces X and Y is the isomorphism constant, de ned as the in mum of kT k T ?1 as T runs over all invertible operators from X onto Y . The closed unit ball of X is denoted by Ball (X). \Local theory" is loosely de ned as the study of properties of in nite dimensional spaces which depend only on their nite dimensional spaces, as well as the study of numerical parameters associated with nite dimensional spaces. Basic for our study and most other investigations in local theory is the fact (see T-J2, p. 54]) that the distance from an n-dimensional space to `n is at 2 most pn. One proves this by using the following consequence of F. John's theorem ( T-J2, p. 123]): If X is n-dimensional and E is the ellipsoid of minimal volume 1 containing Ball (X), then n? 2 E Ball (X). This statement perhaps should be elaborated: Since dim(X) < 1, we can regard X as IRn or C n with some norm. Among all norm-increasing operators u from `n into X, there is by compactness 2 one which minimizes the volume of u(Ball (`n )); the distance assertion says that 2 pn. Alternatively, if one chooses from among all norm one operators from u?1 X into `n one which maximizes the volume of the image of Ball (X), then the norm 2 of the inverse of this operator is at most pn. If complex `n is considered as a real 2 space, then it is isometrically isomorphic to real `2n. Thus the distance statement 2 for complex spaces says that a complex space of dimension n, when considered as a real space of dimension 2n, has (real) distance to (real) `2n at most pn. 2 Actually, we need more than just the distance consequence of John's theorem. The theorem itself T-J2, p. 122] says that if E is the ellipsoid of minimal volume containing Ball (X), then there exist points of contact y1 ; : : :; ym between the unit sphere of X and the boundary of E , and there existPm real numbers 1 ; : : :; m positive summing to dimX so that for each x in X, x = i=1 i hx; yiiyi , where \h ; i" is the scalar product which generates the ellipsoid E . The existence of many contact points between Ball (X) and E is important for the proof of Theorem 3.3. The dual concept to minimal volume ellipsoid is maximal volume ellipsoid. More precisely, an n-dimensional space can be regarded as IRn or C n under some norm k k in such a way that Ball (X) E , where E is the usual Euclidean ball and is also the ellipsoid of minimal volume containing Ball (X). Then X is naturally represented as IRn or C n under some norm, and the action of X on X is given by the usual inner product. Then E is the ellipsoid of maximal volume contained in Ball (X ). John's theorem gives many points of contact between Ball (X) and the boundary of the ellipsoid of minimal volume containing Ball (X), and many points of contact between the boundary of Ball (X) and the ellipsoid of maximal volume contained in Ball (X). It is a nuisance that these two ellipsoids are not generally homothetic (two ellipsoids are homothetic if one of them is a multiple of the other); however, the situation is better when X has the 2-summing property:

The only way that (1.1) and (1.2) are true is if 1 = 2 = = n = 1=pn. But this implies that E2 = E1=pn. Remark 1.2. B. Maurey has proved a far reaching generalization of Lemma 1.1; namely, that if a space X does not have a unique (up to homothety) distance ellipsoid, then there is a subspace which has the same distance to a Hilbert space as the whole space and which has a unique distance ellipsoid. This implies an unpublished result due to Tomczak-Jaegermann which is stronger than Lemma 1.1; namely, that when the distance is maximal, the minimal and maximal volume ellipsoids must be homothetic. Basic facts about 2-summing operators, and, more generally, p-summing operators, can be found in LT1] and T-J2]. The 2-summing norm 2(T) of an operator 1=2 P from a space X to a space Y is de ned to be the supremum of n kTUei k2 1 where the sup is over all norm one operators U from `n , n = 1; 2; : : :, into X and 2 fei gn=1 is the unit vector basis for `n . When Y is a Hilbert space, this reduces i 2 to the de nition given in the rst paragraph of the introduction, and when X is also a Hilbert space, 2(T) is the Hilbert-Schmidt normP T. Note that if U is an of operator from `n to a subspace X of `1 , then kU k2 = n=1 jUei j2 the absolute 2 i value is interpreted coordinatewise in `1 ]. So if T goes from X into a space Y , 2 (T) can be de ned intrinsically by2

(T)2 = supf

n X i=1

kTxik2 :

n X i=1

jxij2

1; xi 2 X; n = 1; 2; 3; :::g;

BANACH SPACES WITH THE 2-SUMMING PROPERTY

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but when Y is an N-dimensional Hilbert space, the \sup" is already achieved for n = N. (Not relevant for this paper but worth noting is that when Y is a general N-dimensional space, the \sup" is achieved for n N 2 FLM], T-J2, p. 141] and p is estimated up to the multiplicative constant 2 for n = N T-J1], T-J2, p. 143].) It is easy to see that 2 is a complete norm on the space of all 2-summing operators from X to Y and that 2 has the ideal property; that is, for any de ned composition T1T2 T3 of operators, 2(T1 T2 T3 ) kT1 k 2(T2 ) kT3 k. The typical 2summing operator is the formal identity mapping I1;2 from L1 ( ; ) to L2 ( ; ) 1 when is a nite measure. In this case one gets easily that 2 (I1;2 ) = ( ) 2 . That such operators are typical is a consequence of the Pietsch factorization theorem ( LT, p. 64], T-J2, p. 47]), which says that if the space X is isometrically included in a C(K) space, and T : X ! Y is 2-summing, then there is a probability measure on K and an operator S from L2 (K; ) into Y so that T is the restriction of SI1;2 to X and kS k = 2(T). That is, there is a probability measure on K so that for each x in X, (1:3)

kTxk2

2 jx (x)j2 d (x ): 2 (T)

Z

Of course, the converse to the Pietsch factorization theorem follows from the ideal property for 2-summing operators. The qualitative version of Dvoretzky's theorem T-J2, p. 26] says that every in nite dimensional space X contains for every n and > 0 a subspace whose distance to `n is less than 1 + . In fact, for a xed n and , the same conclusion is 2 true if dim(X) N(n; ), and the known estimates for N(n; ) are rather good. Here we mention some simple results about spaces which have the 2-summing property, present some motivating examples and then nd a characterization of spaces with that property. Let us say that X satis es the k-dimensional 2-summing property if 2(T) = kT k for every operator T from X into `k . Thus every space has 2 the 1-dimensional 2-summing property, and X has the 2-summing property if X has the k-dimensional 2-summing property for every positive integer k. We introduce this de nition because our techniques suggest that a space with the 2-dimensional 2-summing property has the 2-summing property, but we cannot prove this even in the case of real scalars. Throughout this section the scalars can be either IR or C unless explicitly stated otherwise.(a) If X has the 2-dimensional 2-summing property, then X is nite dimensional. (b) If X has the k-dimensional 2-summing property for some k, then so does X . (c) If X has the 2-summing property, then X is a maximal distance space.1

2. General results.

Proposition 2.1.

property for some integer m. In fact, in the real case, m can be taken to be 3 (Example 2.3), while in the complex case, m = 4 su ces (remark after example 2.3). Alternatively, one can check that a quotient mapping from `1 onto `2 has 2

Proof: For (a), we use the fact that `m fails the 2-dimensional 2-summing

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A. ARIAS, T. FIGIEL, W. B. JOHNSON AND G. SCHECHTMAN

2-summing norm larger than one, which implies that `m fails the 2-dimensional 1 2-summing property if m is su ciently large. So x a norm one operator u from `m into `2 for which 2 (u) > 1. By Dvoretzky's theorem, `2 is almost a quotient 1 2 2 of every in nite dimensional space, so if dimX is su ciently large, then there is an operator Q from X into `2 with Ball (`2 ) Q Ball (X)] but kQk < 2 (u). Pick 2 2 z1 ; : : :; zm in Ball (X) with Qzi = uei for i = 1; 2; : : :; m and de ne T from `m into 1 X by Tei = zi , i = 1; 2; : : :; m. Then u = QT and 2(u) 2 (Q) but kQk < 2(u). For (b), assume that X has the k-dimensional 2-summing property and let T be any norm one operator from X into `k . It is enough to show that 2 (Tu) 1 2 when u is a norm one operator from `k into X . This brings us back to the familiar 2 setting of Hilbert-Schmidt operators: kT k 2(u ) = kT kku k = 1; 2 (Tu) = 2 (u T ) the last equality following from the hypothesis that X has the k-dimensional 2summing property and the fact that, by (a), X is re exive. p For (c), let T : X ! `n be such that kT kkT ?1k = d(X; `n); then n = 2 2 pn. Therefore, d(X; `n) = ?1 ?1 2 (T ?1 2 p2n. T) kT k 2(T) = kT kkT k = d(X; `n ) Remark. To make the proof of (b) as simple as possible, we used (a) to reduce to the case of re exive spaces. Actually, it is well-known that if 2 (T) C kT k for every operator from X into a Hilbert space H, then X has the same property. Indeed, it is easy to see that it is enough to consider nite rank operators from X into H and then use local re exivity (see LT, p. 33]) and a weak approximation argument. Example 2.2. `2 has the 2-summing property. 1 Proof: Let u : `2 ! `2 , kuk = 1. We can assume that 1 2 b u= a d 0 and that (jaj + jbj)2 + jdj2 1; or equivalently, jaj2 + jbj2 + jdj2 + 2jabj 1. For x = (c1 ; c2) 2 `2 we have that 1 2 kuxk =jac1j2 + (jbj2 + jdj2)jc2j2 + 2<(ac1 bc2) (2:1) jac1j2 + (jbj2 + jdj2)jc2j2 + jajjbj(jc1j2 + jc2j2 ): Set = jaj2 + jabj, so that 1 ? (jbj2 + jdj2) + jabj. Thus (2.1) gives (2:2) kuxk2 jc1j2 + jc2 j2(1 ? ): Then since (2.2) is in the form of (1.3) with constant 1, we get that 2(u) 1. Remark. At least in the complex case, Proposition 2.2 follows from the fact that (`2 ) = 1, but we thought it desirable to give a direct proof. Another proof 1 is given in BM]. Proposition 2.6 provides a useful criterion for determining whether a space has the 2-summing property. All of the intuition behind Proposition 2.6 is already contained in Example 2.3:

We denote X by X1 when considered as a subspace of L3 and by X2 when 1 X considered as a subspace of L3 . Also denote by I1;2 the restriction to X of the 2 identity I1;2 from L3 to L3 (we use the standard convention Ln = Ln( ) where 1 2 p p 1 is the probability space assigning mass n to every point). X X We claim that kI1;2 k < 1 and that 2(I1;2 ) = 1. For every kxk1 = 1, kI1;2xk2 1 and we have equality if and only if jxj is at; i.e., x is an extreme point of the unit ball of L3 . Then we verify that 1 X kI1;2 k < 1 by checking that X does not contain any one of those vectors. For the second one, de ne v : `2 ! X1 by vei = xi for i = 1; 2. Then notice that 2 kvk2 = k jx1j2 + jx2j2k1 = 1, where jx1j2 + jx2j2 is taken coordinatewise in L3 , 1 X X 2 = kx1k2 + kx2k2 = 1. The equality follows, since and 2(I1;2 )2 2 (I1;2 v) 2 2 X 2 (I1;2 ) 2 (I1;2 ) = 1. We have thus proved that X does not have the 2dimensional 2-summing property. To conclude, de ne u : L3 ! X2 by u = PI1;2, where P is the orthogonal 1 projection from L3 onto X2 . We claim that kuk < 1 and that 2(u) = 1. 2 If kxk1 = 1, then kI1;2 xk2 = 1 i x is at, and kPxk2 = kxk2 i x 2 X. Since these conditions are mutually exclusive we conclude that kuk < 1. But 1 = kP k2 2(I1;2 )2 2(u)2 2 (uv)2 = kPx1k2 + kPx2k2 = 1. 2 2 1 1 1 The proof for complex `4 is similar: Let x1 = (1; 0; p2 ; p2 ), x2 = (1; 0; pi2 ; p2 ) 1 and X = span fx1; x2g. It is easily checked that X does not contain any at vectors and that jx1j2 + jx2j2 1 coordinatewise. Remark. We shall see in section 4 that complex `3 has the 2-summingproperty. 1 Proposition 2.4. Let X be an n-dimensional subspace of C(K), K compact; u : X ! `k a map satisfying 2(u) = 1 and v : `k ! X satisfying kvk = 1 and 2 2 2 (u) = 2 (uv). Pietsch's factorization theorem gives the following diagram for some probability on K and some norm one operator : X2 ! `k : 2I1;2 C(K) ???! L2 (K; ) i" #P X I1;2 v k ?! `2 X1 ???! X2 u& . `k 2 k ). Then is an isometry on Y2 . Let Y = v(`2 Proof: We have

1 = 2(uv) =2

k X j =1

kuvej k =2

k X j =1

k I1;2 vej k

2

k X j =1

kI1;2vej k2 k k2

2

(I1;2 )2 = 1;

so k I1;2 vej k = kI1;2 vej k for each 1 j k. Recalling the elementary fact that if S is an operator between Hilbert spaces, then fx : kSxk = kS k kxk g is a linear subspace of the domain of S, we conclude that is an isometry on Y2.

Proof: It is clear that the left hand side dominates the right one. To prove the other inequality let u : X1 ! `k be such that 2 (u) = 1. Then nd v : `k ! X1 2 2 such that kvk = 1 and 2(uv) = 1. Let Q be the orthogonal projection from `k onto 2 uv(`k ) = (Y2 ) (with the notation of Proposition 2.4), and P be the orthogonal 2 X projection from X2 onto Y2. Notice that Qu = PI1;2: Since is an isometry X ) = 2(Qu) = 1 and that kPI X k = kQuk kuk. on Y2 we have that 2(PI1;2 1;2 Therefore 1 2 (u) X kPI1;2k kuk :Proposition 2.6 has a nice form when X is 2-dimensional because then we do not need to take the orthogonal projection on X2 . Indeed, if P has rank one then it is X X clear that 2(PI1;2) = kPI1;2k. If is a probability measure on K with support X K0 , then kI1;2k < 1 i Ball(X) does not contain any \ at" vector on K0; i.e., whenever x 2 X and kxk = 1, then we have that jxjjK0 6 1. On the other hand, X 2 2 1 on K and 2 (I1;2 ) = 1 i there exist vectors x1 ; x2 in X such that jx1 j + jx2 j 2 2 jx1j + jx2j 1 on K0 . To see why the second statement is true, nd v : `2 ! X1 2 X satisfying kvk = 1 and 2 (I1;2 v) = 1. Then let xi = vei for i = 1; 2 and the result is easily checked for these vectors. Since every closed subset of a compact metric space is the support of some probability measure, this discussion proves:

where P (K) consists of all the probability measures on K , and I1;2 is the canonical identity from C(K) to L2 (K; ).

BANACH SPACES WITH THE 2-SUMMING PROPERTY

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If X C(K) contains a vector jxj 1, then for every probability measure on X K we have kI1;2xk = kxk; hence, kI1;2k = 1 and we have Corollary 2.8. Let X be a 2-dimensional subspace of C(K), K compact. If X contains a vector x, jxj 1 on K then X has the 2-summing property. This applies immediately to `2 (both real and complex) and also to `2 (again 1 1 real and complex) if embedded in a canonical way. It also implies that there are a continuum of pairwise nonisometric two dimensional complex spaces which have the 2-summing property. We shall see later that the real 2-summing property is quite di erent from the complex version. For the moment, take X = 1 1 span f(1; 0; p2 ); (0; 1; p2 )g inside `3 . We proved in Example 2.3 that real X does 1 not have the 2-summing property. However, complex X has it. To see this, notice 1 1 that (1; 0; p2 )+i(0; 1; p2 ) is \ at" and hence Corollary 2.8 implies the result. The di erence can be explained by saying that it is easier to get \ at" vectors in the complex setting (see Proposition 4.4). Let us return to the discussion following Proposition 2.6. Suppose that X is an n-dimensional subspace of C(K), is a probability measure on K with support K0 K, and P is an orthogonal projection from X2 L2( ) onto a subspace Y2 . X Notice that P2(PI1;2 ) = 1 if and only if there exist vectors x1; x2; : : :; xn in Y n jx j2 1 and each vector I x is in Y . (Keep in mind that with 1K0 1;2 j 2 j =1 j Y2 is relatively L2 ( ){closed in X2 , hence if y 2 Y , z 2 X, and 1K0 y = 1K0 z, then X also z is in Y .) Similarly, PI1;2 = 1 if and only if there exists a single vector x in Y with 1K0 jxj 1. Thus we get a version of Corollary 2.7.a for all nite dimensional spaces: Corollary 2.7.b. Let X be a nite dimensional subspace of C(K), K compact metric. Then X does not have the 2-summing property if and only if there exist Pn 2 1 vectors x1; x2; : : :; xn in X and a closed set K0 K with 1K0 j =1 jxj j such that for every x 2 X with 1K0 jxj 1, we have that 1K0 x is not in span f1K0 x1 ; 1K0 x2; : : :; 1K0 xn g. In the complex case, the 2-summing property is not hereditary, since complex `3 1 has the 2-summing property but `2 is the only two dimensional subspace of it which 1 has the 2-summing property (see Theorem 4.2 and Proposition 5.7.) Nevertheless: Proposition 2.9. Let X be a subspace of `N which has the 2-summing property. 1 Then every subspace of X has the 2-summing property. Proof: Assume that X has a subspace which fails the 2-summing property. Write K = f1; 2; : : :; N g. Since `N = C(K), we can apply Corollary 2.7.b. There 1 exists aPn K0 K for which we can nd vectors x1; x2; : : :; xn in X with subset 2 1K0 1 such that no norm one vector in Y span fx1; : : :; xng j =1 jxj j is unimodular on K0 . We can also assume that K0 is maximal with respect to

Corollary 2.7.a. Let X be a 2-dimensional subspace of C(K), K compact metric. Then X does not have the 2-summing property if and only if there exist vectors x1; x2 in X and a closed set K0 K with 1K0 jx1j2 + jx2j2 1 such that for every x 2 X with kxk1 = 1, we have that jxjjK0 6 1.

10Pn

A. ARIAS, T. FIGIEL, W. B. JOHNSON AND G. SCHECHTMANP

this property; in particular, n=1 jxj j2 is strictly less than one o K0 and hence j 2 j =1 1 K0 jxj j < 1 ? for some > 0. On the other hand, since X has the 2-summing property, there exists a vector y 2 Y which is unimodular on K0 (and whose restriction to K0 agrees with the restriction to K0 of some unit vector in X). Evidently kyk > 1. Thus there exists P 1 2 1 > > 0 so that z = z j 1 yj2 + n=1 j(1 ? ) 2 xj j2 satis es k1 K0 z k1 = 1. j But then z 1, a square function of a system from Y , is unimodular on a set which properly contains K0 ; this contradicts the maximality of K0 . For any real Banach space F, let FC denote the linear space F F with complex structure de ned by means of the formula (a+bi)(f1 f2 ) = (af1 ? bf2 ) (af2 +bf1 ). There is a natural topology on FC , namely FC is homeomorphic with the direct sum of two real Banach spaces F F. In two special cases we shall de ne a norm on FC which will make it a complex Banach space. First, if E is a linear subspace of some real `k , we endow EC with the norm induced from complex `k by means of 1 1 the obvious embedding. Secondly, if E = H is a Hilbert space, then HC is normed by means of the formula kh1 h2k = (kh1 k2 + kh2k2)1=2 . These two de nitions are consistent, because H is isometric to a subspace of real `k only if dimIR H 1. 1 Now, if T : F ! G is a linear operator, we de ne TC : FC ! GC by the formula TC (f1 f2 ) = (Tf1 Tf2 ). Proposition 2.10. Let E be a subspace of real `k and S : E ! `2 be a real-linear 1 2 mapping from E into a 2-dimensional real Hilbert space. Then2

(SC ) = 2 (S) = kSC k :

Proof: From the discussion in Section 1 we see that there are vectors x; y in E such that jxj2 + jyj2 1 (interpreted coordinatewise) and2

linearly independent, since this is true of f f1 ; f2; f3g, so the cardinality of K0 is at least three. But ffi jL g3=1 is linearly independent if L is any three (or four) i element subset of K. To complete the proof, just recall that no norm one linear operator from `3 2 into `2 achieves its norm at three linearly independent vectors since, e. g., the set 2 fx 2 `3 : k xk = k k kxkg is a subspace of `3 . 2 2 Having treated the \worst" case of `n , it is easy to formulate a version of Propo1 sition 3.1 for general spaces. Corollary 3.2. Let X be a real space. (a) If T is a norm one operator from X into `2 such that for some z1 ; z2; z3 in 2 Ball (X), Tz1 ; Tz2 ; Tz3 have norm one and every two of them are linearly independent, then 2(T) > 1. (b) If there exist u : `2 ! X and x1 ; x2; x3 2 Ball (`2 ) such that kuk = 1 = kuxi k = 1 2 2 for i = 1; 2; 3 and every two of x1 ; x2; x3 are linearly independent, then X does not have the 2-dimensional 2-summing property. Proof: For (a), de ne w : `3 ! X by wei = zi ; i 3. Then Tw satis es the 1 hypothesis of Proposition 3.1. Therefore 2(Tw) > 1; and since kwk = 1 we have that 2(T) > 1. For (b) it is enough to prove that X does not have the 2-dimensional 2-summing property. If xi 2 Ball (X ), i = 1; 2; 3 satisfy hxi ; uxii = 1, then 1 ku xi k hu xi ; xii = hxi ; uxii = 1; therefore, u xi = xi for i = 1; 2; 3 and the previous part gives us that 2(u ) > 1. We are now ready for the main result of this section: Theorem 3.3. If X is a real space of dimension at least three, then X does not have the 2-summing property. Consequently, the only real spaces which have the 2-summing property are IR and `2 . 1 Proof: The \consequently" follows from the rst statement and Proposition 2.1(c) because in the real case `2 is the only 2-dimensional maximal distance space. 1 This is an unpublished result of Davis and the second author; for a proof see Lewis' paper L]. So assume that X is IRn , n 3, under some norm and has the 2-summing property. We can also assume that the usual Euclidean ball E is the ellipsoid of minimal volume containing Ball (X), and we use j j to denote the Euclidean norm. P By John's theorem, there exist 1 ; ; m > 0 such that m i = n and i=1 y1 ; ; ym 2 X outside contact Pm (i.e., kyi k = jyi j = 1 for i = 1; 2; : : :; m) such points pn Ball (X), in that every x 2 X satis es x = i=1 ihx; yi iyi . Recall that E = fact, by Lemma 1.1, E =pn is the ellipsoid of maximal volume contained in Ball (X). If x is an inside contact point (i.e., kxk = 1 and jxj = 1=pn), Milman and Wolfson MW] proved that (3:1) 1 jhx; yiij = n for every i = 1; 2; : : :; m:

BANACH SPACES WITH THE 2-SUMMING PROPERTY

To see this, observe that fz 2 X : hz; xi = 1=ng supports E = n at the inside contact point x, hence{draw a picture{the norm of h ; xi in X is 1=n. Thusm m 1 = hx; xi = X hx; y i2 X kh ; xik2 ky k2 = 1 : i i i X i n n i=1 i=11 This implies that jhx; yiij = n for i = 1; 2; : : :; m and proves (3.1). In other words, the norm one (in X ) functional h ; nxi norms all of the yi 's as well as x. This implies that both convfyi : hyi ; nxi = 1g and convfyi : hyi ; nxi = ?1g are subsets of the unit sphere of X. Now we know that X has maximal distance, hence at least one inside contact point exists, whence Ball (X) has at least two \ at" faces. The next step is to observe that we can nd n linearly independent outside contact points y1 ; ; yn and n linearly independent inside contact points x1; ; xn satisfying (3.1). This will give us enough faces on Ball (X) so that a 2-dimensional section will be a hexagon and we can apply Corollary 3.2 (b). Now the John representation of the identity gives the existence of the outside contact points, and since E =pn is the ellipsoid of maximal volume contained in Ball (X), another application of John's theorem gives the inside contact points. So let y1 ; ; yn be linearly independent outside contact points, and x1; ; xn linearly independent inside contact points satisfying (3.1), such that for the rst three of them we have,

p

13

hnx1; y1 i = 1 hnx1; y2 i = 1 hnx1; y3 i = 1 hnx2; y1 i = 1 hnx2; y2 i = 1 hnx2; y3 i = ?1 hnx3; y1 i = 1 hnx3; y2 i = ?1 hnx3; y3i = 1:(We are allowed to change signs and renumber the contact points). Let v denote the 3 linear map from span fy1 ; y2; y3 g into l1 which takes yi to ei . Then kvyk1 = kyk P3 if y = 1 iyi and either 1 2 0 or 1 3 0, hence the restriction of v to F = span f y1 +y2 ; y1 +y3 g is an isometry. But then v Ball (F)] is a regular hexagon, 2 2 which implies that the maximum volume ellipsoid for Ball (F) touches the unit sphere of F at six points. We nish by applying Corollary 3.2 (b). Lewis L] proved that every real maximal distance space of dimension at least two contains a subspace isometrically isomorphic to `2 ; that is, a subspace whose 1 unit ball is a parallelogram. In view of Remark 1.2, the proof of Theorem 3.3 yields: Corollary 3.4. If X is a real maximal distance space of dimension at least three, then X has a subspace whose unit ball is a regular hexagon. As was mentioned in the introduction, the study of the 2-summing property in the complex case is a priori more complicated than in the real case even for two dimensional spaces simply because in the real case there is only one 2-dimensional maximal distance space, while in the complex space there are at least two, `2 14. The complex case.

14

A. ARIAS, T. FIGIEL, W. B. JOHNSON AND G. SCHECHTMAN

and its dual. In fact, it is not di cult to construct other complex 2-dimensional maximal distance spaces without using Corollary 2.8. One way is to use John's representation theorem; this was done independently by Gowers and TomczakJaegermann both unpublished] a couple of years ago in order to construct real 4-dimensional maximal distance spaces whose unit ball is not strictly convex; this approach also yields maximal distance 2-dimensional complex spaces di erent from `2 and `2 . However, a simpler way of seeing that there are many maximal distance 1 1 2-dimensional complex spaces is via Proposition 4.1: Proposition 4.1. Suppose that X is an n-dimensional complex space which, as a real space, contains a real n-dimensional subspace which has maximal distance to real `n . Then X has maximal distance to complex `n . 2 2 Proof: Let Y be a real n-dimensional subspace of X which has maximal distance to real `n , and suppose that T is any complex linear isomorphismfrom X to complex 2 `n . But considered as a real space, complex `n is just real `2n, and so the restriction 2 2 2 of T to Y is a real linear isomorphism from Y to a real n-dimensional Hilbert space, hence by the assumption on Y , pn; kT k T ?1 TjY (TjY )?1 so the desired conclusion follows. Proposition 4.1 makes it easy to construct in an elementary manner 2-dimensional complex maximal distance spaces which are not isometric to either `2 or `2. For 1 1 example, in complex `3 , let x = (1; 0; a) and y = (0; 1; b) with ja + bj _ ja ? bj 1 1 but jaj + jbj > 1, and set X = span fx; yg. In Proposition 4.4 we prove that every 2-dimensional subspace of complex `3 is a maximal distance space, by proving that 1 they all have the 2-summing property. However, in the complex setting, the 2-summing property is not restricted to two dimensional spaces. In fact, we do not know a good bound for the dimension of complex spaces which have the 2-summing property, although we suspect that dimension three is the limit. In dimension three itself, we know of only two examples, `3 and its dual. 1 We can prove Theorem 4.2 using Proposition 2.6 and The proof of Proposition 4.4 (see Remark 4.6). The proof we give is of independent interest. Theorem 4.2. Complex `3 has the 2-summing property. Hence also `3 has the 1 1 2-summing property. Proof: The proof reduces the theorem to the following calculus lemma, which we prove after giving the reduction: Lemma 4.3. Fix arbitrary complex numbers 1, 2, 3 and de ne a function f f(z1 ; z2) = j1 + 1 z1 + 2 z2 + 3 z1 z2j + j 3j 1 ? jz1j2 1 ? jz2 j2: Then the maximum of f is attained at some point on the two dimensional torus; that is, when jz1 j = jz2j = 1. Reduction to Lemma 4.3: Let K = f1g T T, where Tis the unit circle, and regard `3 as the subspace of C(K) spanned by the coordinate projections f0 ; f1; f2 . 1on the bidisk byp p

BANACH SPACES WITH THE 2-SUMMING PROPERTY

15

Let u be a norm one operator from `3 into `2 (complex scalars). We want to show 1 that 2(u) = 1. That is, we want a probability on K so that for each x 2 `3 , 1

kuxk2

Z

jx(k)j2 d (k):

Notice that if we add to the ufj 's mutually orthogonal vectors which are also orthogonal to the range of u, the 2-norm of the resulting operator can only increase. Thus we assume, without loss of generality, that kufj k = 1 for j = 0; 1; 2, and that uf0 = 0 ; uf1 =1 0

Choose so that ei 1 3 is purely imaginary; so <( 1 3 z2) does not change if you add a real multiple of ei to z2 . Assume rst that 2 6= 0. Then, for any > 0, by adding either ei or ? ei to z2 you increase the rst term; this means that at the maximum jz2j = 1. The case where 2 = 0 can be handled in a similar way. Remarks. 1. A similar argument reduces the problem of whether `4 has the 1 2-summing property to a calculus problem; however, in the remark after Example 2.3 we give a simple direct argument that `4 fails the 2-summing property. 1 2. Bagchi and Misra BM] give a di erent reduction of Theorem 4.2 to a variation of Lemma 4.3. Their argument may be more appealing to operator theorists. 3. The following Proposition is a consequence of Proposition 2.9 and Theorem 4.2. We indicate a shorter argument. Proposition 4.4. Every two dimensional subspace of `3 has the 2-summing prop1erty.

Proof: Assume that X `1 is two dimensional. Applying `1 isometries we assume that X has a basis of the form f(1; 0; a ); (0; 1; a )g where jai j 1 for i = 1; 2. If ja j + ja j 1 then X is isometric to `1 and by Example 2.2 it has the 2-summing property. So assume that ja j + ja j > 1. It is easy to nd , 0 < 2 such that ja + ei a j = 1. Hence, (1; 0; a ) + ei (0; 1; a ) is \ at" and the result3 3 1 2 1 2 2 1 2 1 2 1 2

follows from Corollary 2.8. It is interesting to notice that Proposition 4.4 is equivalent to the following calculus formulation. Proposition 4.5. Given any complex numbers c1; c2; c3 and d1; d2; d3 with jcj j2 + jdj j2 1 for j = 1; 2; 3, suppose that ; satisfy

(4:1)

j j 2j j 2 + j j 2 j j 2

j =1;2;3

max j cj + dj j2 8 ; 2 C :

Then j j2 + j j2

1. To see the equivalence of Propositions 4.4 and 4.5, let X be a 2-dimensional u v subspace of `3 and X ?!`2 a norm one operator. Choose `2 ?!X of norm one 1 2 2 so that 2 (u) = 2(uv). We can assume that uv is diagonal; say, uv(e1 ) = e1 and uv(e2 ) = e2 . For j = 1; 2 set xj = v(ej ) and write x1 = (c1 ; c2; c3), x2 = (d1; d2; d3). So 2(u)2 = j j2 + j j2, while kvk2 = j =1;2;3jcj j2 + jdj j2 = 1. The max implication Proposition 4.5 ) Proposition 4.4 follows by noticing that u having norm at most one is equivalent to the inequality (4.1). Similar considerations yield the easier reverse implication. We do not see a really simple proof of the calculus reformulation of Proposition 4.4 without using Pietsch's factorization theorem. However, a similar reduction of the weaker statement that every 2-dimensional subspace of `3 has maximaldistance 1 to `2 produces a calculus statement which is very easy to prove. Indeed, given a 2 u 2-dimensional subspace X of `3 , we can choose norm one operators X ?!`2 and 1 2 v X so that duv = I 2 , where d is the distance from X to `2. We can choose `2 ?! `2 2 2 the orthonormal basis e1 ; e2 so that 1 = kvk = kve1 k and de ne x1; x2, the ci 's,and

BANACH SPACES WITH THE 2-SUMMING PROPERTY

17

the di's as in the discussion above. Since kx1k = 1, we can assume, without loss 1 of generality, that jc1p= 1. (4.1) holds with = = d , and we want to see that j 2 2 this implies that d 2; i.e., that + 1. So we only need to get and of modulus one to make the right side of (4.1) one. Since jc1j = 1, d1 = 0, any such choice makes j c1 + d1 j = 1. Choose to make c2 0; then j c2 + d2 j 1 as long as d2 has nonpositive real part, which happens as long as is on a certain closed semicircle. Similarly, j c3 + d3j 1 as long as is on another closed semicircle. Since any two closed semicircles of the unit circle intersect, the desired choice of and can be made. Remark 4.6. 1. We can use Propositions 2.6 and 4.4 to prove that `3 has 1 the 2-summing property. Use the notation of Proposition 2.4 and let P be an orthogonal projection on L3 ( ). If the rank is one, there is nothing to prove. If the 2 rank is three then we clearly have that kI1;2k = 1, and if the rank is two, then the proof of Proposition 4.4 implies the result. 2. It is natural to ask if the only subspaces of complex L1 with the 2-summing property are `2 and `3 . The answer is yes because a subspace of L1 of maximal 1 1 distance is already an `k space. We prove this in the appendix, Proposition 5.7. 15. Appendix.

In this section we present some related results. Proposition 5.1. Every subspace of complex `3 is the complexi cation of a sub1 space of real `3 1 The proof of Proposition 5.1 follows easily from the next two lemmas. Recall that a vector in `k is said to be at if all of its coordinates are unimodular. 1 Lemma 5.2. Suppose that the subspace X of complex `3 is not linearly isomet1 2 ric to l1 . Then X contains two linearly independent at vectors, say f1 and f2 . Moreover, each at vector in X is of the form fj , where j 2 f1; 2g and j j = 1. Proof: Applying `3 isometries, we may assume that X is spanned by two 1 vectors of the form x = (1; 0; a) and y = (0; 1; b) where a; b 2 C with jaj; jbj 1. Put w = x ? y. For w to be at one needs that j j = 1 and ja ? bj = 1. Observe 2 that, since X is not linearly isometric to l1 , we have jaj + jbj > 1;, in particular ab 6= 0. Thus 2 C should belong to the intersection of the unit circle fz : jz j = 1g and the circle fz : jz ? a=bj = 1=jbjg, hence there are at most two solutions for . Thus it will su ce to check that the two circles have a point in common and that they are not tangent at that point. Since the second circle has a bigger radius, this amounts to verifying the strict inequalities 1 ?1< a ?0 < 1+ 1 : b b b These inequalities are obvious, because we have jaj + jbj > 1, jaj 1 and jbj > 0. Lemma 5.3. Suppose that the 2-dimensional subspace X of complex `k is spanned 1 by two linearly independent vectors y; z such that jyj = jz j. Then there is a linear

18

A. ARIAS, T. FIGIEL, W. B. JOHNSON AND G. SCHECHTMAN

k isometry of l1 such that y = z . In particular, (X) is spanned by two vectors v; w, all of whose coordinates are real and which satisfy (jvj2 + jwj2)1=2 = jyj.

Proof. Write y = (y1; y2; ; yk) and z = (z1; z2; ; zk ). For j = 1; 2; ; k, let j be a complex number with j j j = 1 such that j yj = j zj . Such numbers obviously exist, we may also impose the condition < j yj 0. Now the isometry can be de ned by the formula (x1 ; x2; ; xk ) = ( 1 x1; 2x2; ; kxk ). Clearly, the vectors v = 1 ( y + z) and w = 21i ( y ? z) have the required property. 2 Propositions 2.9, 2.10, and 5.1 suggest an alternate method for proving Proposition 4.4 since they combine to take care of the case where the operator achieves its norm at two \ at" vectors: Lemma 5.4. Let X be a two-dimensional subspace of complex `3 and T a complex1 linear operator from X into a Hilbert space such that kTxk = kTyk, where x; y are linearly independent vectors in X for which jxj = jyj. Then 2(T) = kT k. Proof: In view of Lemma 5.3 we can assume that there are vectors v, w in X all of whose coordinates are real for which jvj2 + jwj2 = jxj2, x = v + iw, and y = v ? iw. Thus if we let E be the collection of real-linear combinations of fv; wg, we can regard X as the complexi cation EC of E. The assumption on fx; yg means that the pair fv; wg satis es condition (iv) in Proposition 2.11, hence condition (i) of Proposition 2.11 says that T is the complexi cation of the restriction of T to E, whence by Proposition 2.10, 2(T) = kT k. The next lemma takes care of the case where the operator achieves its norm at a non- at vector. Lemma 5.5. Suppose that X is a 2-dimensional subspace of complex `3 and the 1 T norm one operator X ?!`2 achieves its norm at a non- at vector x = (x1 ; x2; x3) 2 W V on the unit sphere of X . Then there are norm one operators X ?!`2 and `2 ?!`2 2 1 1 so that T = WV . Consequently, by Example 2.2, 2(T) = 1. Proof: Since the result is trivial if T has rank one, we assume that T has rank two. This implies that two coordinates of x, say, x1 and x2, are unimodular. Indeed, suppose, for example, that jx2j and jx3j are both less than 1 ? . Take y in X with y1 = 0 and 0 < kyk < , so kx yk = 1. But since `2 is strictly convex, kT(x + y)k > kTxk for either = 1 or = ?1.] We may also assume that X contains two vectors, say y; w, such that y = (1; 0; y3) and w = (0; 1; w3). (Otherwise, X is spanned by two vectors with disjoint supports and the conclusion of the Lemma is obvious.) Let be the function de ned for z 2 C by the formula(z) = kT(x1y + zw)k2 : Observe that (z) = kT(x1y) + zTwk2 is a quadratic function of (<z; =z), which at in nity is asymptotically equal to mjz j2, where m = kTwk2 > 0. It follows that there is a number z0 2 C such that (z) = mjz ? z0 j2 + (z0 ) for every z 2 C . It is obvious now that either z0 = 0, so that is constant on the unit circle, or else has a unique local maximum on the unit circle (which must also be the global maximum of on the circle). Note that at z = x2 the function does have a

We are going to show that, for some choice of the parameters, jxj2 +jyj2 is constantly equal to one while for the same choice span fx; yg does not contain a at vector. Once this is proved one concludes the proof as in Example 2.3. jxj2 + jyj2 1 is equivalent to (5:1) 1=2

+

2

=

2

+

2

= j a1 + 'a2 +

a3 j2 + j a1 ? a3 j2

while, if jax + byj then (5:2)

1, we may assume without loss of generality that a = 1 and a3 + b( a1 ? a3 )j:

same or opposite direction. If a2 (2) + (a (1) ? a (3) )2 1 for all permutations, , of the indices 1; 2; 3 for which a (2) > 0 then, assuming as we may that a1 a2; a3 and a1 > 0, it is easily checked that there are ; > 0 with 2 + 2 = 1 for which

Moreover, the sup is attained for ( ; ) proportional to (a2 ; a3). For this choice of ( ; ), j a1 ? a2 j2 + j a1 ? a3j2 = (a1 ? (a2 + a2)1=2 )2 < 1. Choose now = 0; = 1 and 2 3 notice that there is a one parameter family of '; for which j a1 + 'a2 j2 + j a1 ? a3 j2 = 1 but not for all of members of this family do a1 + 'a2 and a1 ? a3 point in the same or opposite directions. Remark. It is easy to adjust the proof above to show that for n > 3 no ndimensional subspace of `n+1 has the 2-summing property. Indeed without loss of 1

BANACH SPACES WITH THE 2-SUMMING PROPERTY

21

generality any such subspace is spanned by n vectors of the form (1; 0; : : :; 0; a1); : : :; P (0; : : :; 0; 1; an) with 0 ai 1. If ai 1 the subspace is isometric to `n which 1 does not have the 2-summing property. Otherwise these n vectors can be blocked to get three vectors which could replace the three vectors with which we started the proof above. We next present a proof that all maximal distance subspaces of L1 are `n spaces. 1 In particular there are no new subspaces of (real or complex) L1 with the 2-summing property. Essentially the same proof shows that all maximal distance subspaces of Lp are `n spaces, 1 p < 1. This fact is not new: it was observed by J. Bourgain p that the case p = 1 follows from FJ]. The case 1 < p < 2 was rst proved in BT]. Komorowski Ko] was the rst to prove the 2 < p < 1 case. The proof here is very similar to Komorowski's but includes also the 1 p < 2 case. Proposition 5.7. Let 11 1p < 1 and let X be n-dimensional subspace of an Lp( ) space with d(X; `n ) = nj p ? 2 j , then X is isometric to `n . p 2 Proof. By Lewis' theorem L] we may assume that is a probability measure and that X has a basis x1; x2; : : :; xn satisfying (5:3) and (5:4) Then, for 1 p 2,n n n 1 (X ja j2 )1=2 = (Z j X a x j2 d )1=2 (Z j X a x jp d )1=p i i i i i n1=2 i=1 i=1 i=1 n X i=1Z

i=1 This shows that,1if T is the map sending the xi 's to an orthonormal basis, then 2 jjT jj jjT ?1jj nj p ? 1 j . It follows that there are a1 ; a2; : : :; an for which equality is

achieved in both (5.5) and (5.6). If 2 < p < 1, then we get similarly thati=1

n n 1 (X ja j2)1=2 (Z j X a x jp d )1=p i i i n1=2 i=1

and (5:50) (5:60) ( jZ

n X i=1

X X p?2 ai xijp d )1=p ( j ai xij2d )1=p sup j aixi j p

Z

n

n

( j

Z

n X

i=1

1 = n1=p (

i=1 n X i=1

ai xij2d )1=p (

n X i=1

i=1

jaij2) 2p

p?2

jaij2)1=2 :

Again we get that some a1; a2; : : :; an must satisfy (5:50) and (5:60) as equalities. Examining when equalities can occur in (5:5); (5:50); (5:6) and (5:60), we see that P for all p there are a1; a2; : : :; anP that j n=1 ai xi j is a constant on its support, such i A, and the constant must be ( n=1 jaij2)1=2 which we may assume is equal to 1. i Moreover, on A, (x1(t); x2 (t); : : :; xn(t)) must be equal to (t)(a1 ; a2; : : :; an) for some function satisfying j (t)j 1 . Applying a space isometry,Pn may assume that 1 and R we P 1 then n=1 ai xi = A . Note also that (A) = j i=1 ai xij2d = n . We thus get i that we may assume that A 2 X. Since each xi is constant on A, we get that for all f 2 X, fjA is a constant and fjAc also belongs to X. Put Y = ffjAc : f 2 X g;1 1 n then Y X and dimY = n ? 1. Necessarily d(Y; `p ?1) = (n ? 1)j p ? 2 j and continue...